Shortcuts In Reasoning Competitive Exams – Problem Solving

Shortcuts in Reasoning Quantitative Aptitude English

INTRODUCTION

In this chapter you will see some typical problems in which you would be given a series of interlinked information and on the basis of those information you would be expected to reach certain conclusions.

TYPES OF INFORMATIONS IN A GIVEN PROBLEM

Basic Informations(Useful secondary informations):
It is given in first couple of sentences of given data are such that they give you some basic information that is essential to give you general idea of the situation.

Actual Informations
Whatever remains after the basic informations are known as actual information.
While trying to solve a problem one should begin with actual information and useful secondary information should be solve by mind.

Negative Informations
Actual informations having negative sentences are called negative information. A negative information does not inform us anything exactly but it gives a chance to eliminate a possibility.

TYPES OF PROBLEMS

1. Simple problems (based on categorisation)
2. Problems based on arrangement (Linear, circular, rectangular/ square).
3. Problems based on comparison.
4. Problems based on blood relations.
5. Blood relations and profession based problems.
6. Problems based on conditional selection.

SIMPLE PROBLEMS BASED ON CATEGORISATION
Tips to Solve Problems
These type of problems can easily be solved by construction of table.

EXAMPLE
Directions : Read the following information carefully and answer the question that follows:

1. There are six cities L, M, N, O, P and Q.
2. L is not a hill station.
3. M and P are not historical places.
4. O is not an industrial city.
5. L and O are not historical cities.
6. L and M are not alike.

Question: Which two cities are industrial centres ?
Solution:
It can be solved by preparing a table in the manner given below:

(2), (3), (4), (5) are negative informations. Therefore as per such informations. We put ‘X’ (not) mark wherever applicable. As a result the table looks like the one below.

As above table gives definite informations about L,O.

• L is neither a historical place nor a hill station.
• So, it must be an industrial city.
• In the same manner O is neither a historical nor an industrial city.
• So, O must be a hill station.
• Hence, we put mark at the appropriate place which give the table following look:-

Now, as per the condition (6) (L and M are not alike), M can not be an Industrial city.
Also M is not a historical place either. Therefore, it is very obvious that M is a hill station.
Again, in the given problem there is no negative information about N.
Hence, we can assume that N is a hill station as well as a historical place and an industrial city.
Combining if these aspects, the following table will be prepared finally.

Now, after analysing the given question we get the answer:- So, P and Q are two industrial centres.

PROBLEMS BASED ON ARRANGEMENT
In such problems a group of people, objects, etc, may have to be is arranged in a row, or in a circle or any other way.

Linear Arrangement
one row sequence

• When direction of face is not clear, then we take ourself as base and then the diagram will be as follows
  
  
  From the above diagram, it is clear that
  B, C, D, E are right of A but only B is the immediate right of A.
  D, C, B, A are left of E but only D is the immediate left of E.
• When direction of face is towards you, then the diagram will be as follows
  
  

From the above diagram, it is clear that

• B is immediate left of A, C is immediate left of B;
• D is immediate left of C and E is immediate left of D;
• D is immediate right of E; C is immediate right of D;
• B is immediate right ofC; and A is immediate right of B.

Two rows sequence
Let us see 6 persons seating in two rows

From the above diagram, it is clear that

• P is sitting opposite S.
• Q is sitting opposite T.
• R is sitting opposite U.
• P and U are sitting at diagonally opposite positions.
• S and R are sitting diagonally opposite positions.

Note:
Point to be noted that in arrangement problems, the actual information can be classified into 2 categories

• Definite information
  A definite information is one when the place of object/man is definitely mentioned.
• Comparative information
  In such information the place of object/man is not mentioned definitely but only a comparative position is given. In other words the positions of objects/men are given in comparision to another objects/men.

Shortcut Approach

1. Sketch a diagram of empty places
2. Fill up as many empty places as possible using all the definite informations.
3. With the help of comparative information consider all possibilities and select the possibilities which does not violate any condition

EXAMPLE:
Directions: Just read the following information carefully to answer the questions given below it:
Five friends P, Q, R, S, and T are sitting on a bench.

1. Pis sitting next to Q.
2. R is sitting next to S.
3. S is not sitting with T.
4. T is on the last end of the bench.
5. R is on the 2nd position from the right.
6. P is on the right of Q and T.
7. P and R are sitting together.

Question: All what position is P sitting?
Solution:
Here, 4th and 5th sentences constitute definite information:
Comparative informations are: 1st, 2nd, 6th and 7th sentences while 3rd is a negative information.
Now, start with definite information, sketch the following arrangement:-
T _ _R_

Now, this is the time to look for the comparative informations that tell about T and R. Such informations are 2nd, 6th and 7th sentences. Take the 7th and the 1st sentence. If P and R are together and also Q and P are together, then P must be between Q and R. Now the arrangement take the form as:-
T Q P R _
By the virtue of the 2nd sentence:
T Q P R S
So, P is sitting between Q and R.

Circular Arrangement
Circle is the most important case from the exam point of view. Most of the times Circle kind of statements are there in exams.
From the exam point of view, in most cases they give 8 persons sitting in the circle. But before solving the important thing is their’ Sitting Position’.

Step 1.
Knowing NEWS!  N= North, E= East, W=West, S= South

To remember this just remember combination’ North • comes together to each other respectively.

Step 2 :
Picking Left & Right.

If it is mention in the statement that all is 1 facing outside then just do opposite of above like this:
Clock wise = Right & Anti-clock wise ‘
= Left I

Step 3 :
Solving step wise the statement or Following the statement.

Shortcut Approach

• Imagine yourself as one of the persons given in the question.
• Count how many people are mentioned in the question. Then draw a circle with those many lines.

• Imagine yourself at the position shown by the box.
• Now your left hand is the left side and right hand is the right side. Now, if in question it is given, P is second to the right of Q, approach as follows.
  —> Imagine yourself as Q.
  
•  Now, P is second to right of1 Q. The right ofQ is your right side.’ So, place P two places from Ql towards its right.
  

EXAMPLE
Directions:
Study the following information carefully and answer the question given below.

Bunty, Dev, Manav, Kavya, Payal, Qasturba, Wasir and Himmat are sitting around a circle facing at the centre. Manav is to the immediate right of Bunty who is 4th to the right of Kavya. Payal is 2nd to the left of Bunty and is 4th to the right of Wasir. Qasturba is 2nd to the right of Dev who is 2nd to the right of Himmat.

Question: Who is 3rd to the right of Bunty?
Solution:

Now, look at the given question and check that you get the answer.
So, Himmat is 3rd to the right of Bunty.

PROBLEMS BASED ON COMPARISON
In such problems comparison of different objects or persons has to be made. Such comparisions are done on the basis of marks, ages heights, etc.

Method to Solve
If you give a serious look to the problem you will find that such problems are as same as the arrangement problems. Therefore, we have to go like arrangement problem while solving problems based on comparison.

EXAMPLE
Directions : Read the informations given below to answer the given question:

1. 7 students A, B, C, D, E, F and G take a series of tests.
2. No two students obtain the same marks.
3. G always scores more than A.
4. A always scores more than B.
5. Each time either C scores the highest and E gets the least, or alternatively D scores
   the highest and F or B scores the least.

Question: If D is ranked 6th and B is ranked 5 th, which of the following can be true?
Solution:

In this case, we see there is no definite information. Sentence 5 gives a definite information but it is conditional. Still, we draw all the possibilities based on sentence 5.

1. C———–E
2. D———–F
3. D———–B

We see that the two additional informations (3) and (4) are inadequate to reach a definite conclusion. Hence, keeping these in mind. We move on to the given questions.
D is ranked 6th and B is 5th. This does mean that possibilities (2) and
(3) are violated. Hence, possibility (1) must be true. Thus, we have:
C_____B D E
Also by virtue of(3) and (4) we can have only one arrangement for G, A and B which is GAB.
Accordingly, there are two possibilities:
C G F A B D E
C G A F B D E
So, if D is ranked 6th and B is ranked 5th, then f is ranked 3rd or 4th.

PROBLEMS BASED ON BLOOD RELATION
Such problems involves analysis of certain blood relations.

Shortcut Approach

• Vertical/diagoinal lines to represent parent-child,relationships.
• Single/double horizontal line like (<=>) to represent marriages.
• A dashed line (—) for brother and sister relationship.
• ‘+’ sign for male and ‘-’sign for female .For example.
  
  or
  
  or
  

The above diagrams tells us:

1. A and Bare couple; A is the | husband while B is the wife.
2. D is son of A and B while E i is daughter of A and B.
3. D is the brother of E and E is the sister of D.
4. D has a son F.
5. F and G are couple; F is the. husband and G is the wife.
6. F is the grandson of A and B
7. G is the daughter in law of D.
8. E is the aunt (Bua) of F
9. There are 3 males (A, D and F) and 3 females (B, E, G)

EXAMPLE:
Directions : Read the following information carefully and answer the question given below:
There are 6 members in a family. They are M, N, O, P, Q, R are travelling together. N is the son of O but O is not the mother of N. M and O are a married couple. Q is the brother O. P is the daughter of M. R is the brother of N.

Question: How many male members are there in the family?
Solution:
Here, all the sentences are actual information except the first out of these the 2nd and the fifth sentences give information on parent child relationship. We can begin with either of the two. Let us begin with the 6th sentence. Our diagram will be as

As, we do not want to make many diagrams and instead we would prefer to only add to the existing diagsams. Therefore, we should look for sentences that talk of M or P. The 3rd sentence talks about M. Hence, we add this information, that M and O are married couple in our diagram.

Now, the 2nd sentence talks about O. It says that N is the son of O but O is not the mother of N. Obviously, O must be the father of N. This means O is a male and hence M must be a female. Now our diagram takes the form as following:-

Now, we add the two sentences ‘Q is the brother of O’ and ‘R is the brother of N’ and we get the final diagram as below:-
So, there are 4 male members in the family.

PROBLEMS BASED ON BLOOD RELATIONS AND PROFESSION

Such problems are very much similar to the problems related to blood relation. What makes it different is the addition of new data:- the professions of family members. You will get the more clear idea about this type of problem.

EXAMPLE:
Directions : Read the following information carefully and answer the question given below it:

1. A, B, C, D, E and P are members of a family.
2. There are two married couples.
3. B is an engineer and the father ofE
4. P is the grandfather of C and is a lawyer.
5. D is the grandmother of E and is a housewife.
6. There is one engineer, one lawyer, one teacher, one housewife and two students in the family.

Question: Who is the husband of A?
Solution:
Here, (1), (2), and (6) are useful secondary informations. While (3), (4) and (5) are the actual informations. We start with the 3rd sentence because it mentions a parent.
Child relationship its diagram can be made as the following:-

Now, we see that we have ended up with two different component. Then how to resolve this deadlock? The answer is simple: – to resolve it, we make use of the given useful secondary information (USI). “There are two married couple in the family.” Clearly, the two possible pairs are of grandfather, graiidmother and father, mother. Therefore, we combine the two diagrams into the following way.

Point to be noted that the professions of A, E and C are yet unknown. However, with reasonable justification, we may assume that the mother (A) should be the teacher and the two children E and C should be students. But this conclusion can be challenged and has no reason at all.A part from that the sexes of E and C can not be determined. So, B is husband of A.

PROBLEMS BASED ON CONDITIONAL SELECTION

In this type of problems, a group of objects/persons has to be selected from a given larger group, as per the given restrictions. You will get the better idea of such type of problem from the problem given below:-

EXAMPLE:
Directions :
Study the following information carefully and answer the question given below:-
From, amongst 6 boys J, K, L, M, N, and O and 5 girls P, Q, R, S and T, a team of 6 is to be selected under the following conditions:-

1. J and M have to be together.
2. L can not go with S.
3. S and T have to be together.
4. K can not be teamed with N.
5. M cannot go with P.
6. K and R have to be together.
7. L and Q have to be together.

Question: If there be 5 boys in the team, then the lone girl member is——–
Solution:
Make the group of all the pairs that have to be together on one side and the pairs that must not be together on the other side. Next, read each of the questions and treat that as an additional information. Finally, analyse the possibilities and choose the possibilities that satisfies all the conditions.
Let us see the process below:-
Firstly, we can summarise the conditions in the following way:-

Here, number of boys are 5. We see than K and N can never be
together. Therefore, there are only two ways of selecting 5 boys:- JKLMO and JNLMO. But the possiblity is not possible because if K would go then R should also go, and if L goes than Q should also go. Hence, JNLMO is the only possibility in which L’s friend Q would be the lone girl member.

Shortcuts in Reasoning for Competitive Exams – Input and Output

Shortcuts in Reasoning Quantitative Aptitude English

INTRODUCTION

Problems related to input-output are frequently asked questions in various graduate level competitive examinations. They are not very tough stuff but take a good deal of time to be solved or sometimes students do not take attempt to solve them because of time consuming impression of such type of questions. But proper understanding of the subject makes you believe that such problems are not as tough and time consuming as they seem.

CONCEPT OF INPUT- OUTPUT PROBLEMS
In such problems:

• It is imagined that there is some kind of computer/word processing machine.
• An input is given to the computer/ word processing machine
• The computer/word processing machine performs repeated operations as per a certain pattern to give different output in different steps.

TYPES OF PROBLEMS

• Problems of shifting
• Problems of arrangement
• Problems of mathematical operation
• Miscellaneous.

PROBLEM OF SHIFTING

We know that in such type of problems, a word/number processing machine generate output through shifting. Shifting does mean an operation in which words or numbers of a given input give outputs in different steps through shifting their place to different place as per a fixed pattern.

Note: In shifting problems, the previous step of any step can possibly be determined, so we can move in backward or reverse order which is not possible in some of the other type of problems.

Methods to Solve
Lets take an example
Input: Blue Cat Good Other Have Cake
Step 1: Blue Other Good Cat Have Cake
Step 2: Blue Other Have Cat Good Cake
Step 3 : Cake Other Have Cat Good Blue
Step 4: Cake Cat Have Other Good Blue
Step 5 : Cake Cat Good Other Have Blue
Step 6: Blue Cat Good Other Have Cake
Shifting of element can easily be understood by making them equivalent to number like
Blue = 1, Cat = 2, Good = 3, Other = 4, Have = 5, Cake = 6

Input can be written as
1…..2……3……4…..5…..6
Blue Cat Good Other Have Cake
Step-1: 2 and 4 interchanged
Step-2 : 3 and 5 interchanged
Step-3 : 1 and 6 interchanged
Step-4 : 1,2 and 3 are repeated again.

PROBLEMS ON ARRANGEMENTS

Word Arrangement from Left Side:
EXAMPLE:

Here, we start arrangement from the word that comes 1st in the dictionary; then comes the word coming 2nd in the dictionary, then comes the word coming 3rd in the dictionary and so on. In this case, the arrangement start from left side. This is the reason in step I banana comes 1st as it comes 1st in the dictionary. In the 2nd step, orange comes at 3rd place because after the arrangement of step I the next word coming in the dictionary is mango but it get arranged automatically and hence there is no need to arrange it in step II. This is the reason after arranging banana in step I, we directly come to the word orange (coming 3rd in the dictionary) in step II. In the 3rd step, we arrange the word ‘pear’ (coming 4th in the dictionary) and the word tango get arranged automatically.

Word Arrangement from Right:
EXAMPLE:

In this case, the arrangement starts from right side. The word coming 1 st in the dictionary comes at the 1 st position from right. At the 2nd position from right comes the word coming 2nd in the dictionary and the process goes on till the arrangement gets completed. In the above given example, ‘Fame’ is the 1st word coming in the dictionary and hence it comes at the 1 st position from right in the step I. In the step II, the 2nd word coming in the dictionary (Game) comes at the 2nd position from right. Point to be noted that the word coming third in the dictionary will come at the 3rd position from right and this word is ‘Jam’. But ‘Jam’ automatically get arranged as per the given pattern when we arrange the word ‘Game’ in II step. This is the reason why we don’t arrange ‘Jam’ in the third step and jump directly to arrange the word. ‘Name’ that comes 4th in the dictionary. ‘Name’ occupies 4th position from right and the word ‘Shame’ automatically get arranged in the 3rd step. Hence, the word ‘Shame’ does not need to get arranged.

Word Arrangement from the Left-Right Alternate:
EXAMPLE:

Here, the arrangement is made by putting the first word at 1st place, then alphabetically last word at last place, then alphabetically second word at second place from left and the further arrangements goes on in the same manner. In the other words, are positioned from the left and from the right alternately. In the step I the word coming 1 st in the dictionary is ‘ a’ and it takes 1 st position from left. In the step II, the last word coming alphabetically is Sachin and it takes last position (1 st from right). In step III, the word coming 2nd in dictionary is ‘cricket’ that comes at 2nd position from left. In step IV, the word coming 3rd last in the dictionary takes the 3rd position from right. After the step IV, all the words get arranged in alphabetical order. Point to be noted that after step IV, there is no need to arrange the word ‘great’ as it get arranged automatically is step IV.

Arrangement in Increasing or Decreasing Order:
EXAMPLE :

This arrangement gives a clear idea of arrangement of numbers in increasing order. In step I, the smallest number (17) comes at the 1st position from left pushing the remaining to the right. In step II, the 2nd smallest number (18) comes at 2nd position from left pushing the remaining number to the right. In step III, the 4th smallest number (35) takes 4th position from left and the other two numbers 58 and 100 get arranged automatically.

Now, let us see decreasing order arrangement:

The same arrangement can take place from right side (or) in the reverse order as follow:

Number Arrangment from Left-Right Alternate:

Like words left-right alternate arrangement, number arrangement also takes place. The process of this arrangement is exactly the same as the arrangement takes place in case of words. Just see the following cases:

Case I:

Here, the smallest number (10) takes 1 st position from left in step I. In step II the largest number takes the last (1st from right) position. Again in step III the 2nd smallest number (15) comes at the 2nd position from left. In the step IV, the 2nd largest number (100) comes at the 2nd position from right and the remaining number (26 and 35) get arranged automatically.

In case II, the arrangements take place in the same way as the arrangements take place in case I. But the difference here is that case I is a left-right arrangement and case II is the right-left arrangement. In case II, the arrangement starts with the largest number (125) coming at the 1st position from right and this is step I. In step II, the smallest number (10) comes at the 1 st position from left. In step III the 2nd largest number (100) comes at the 2nd position from right. In step III, the third largest number (35) automatically comes at the 3rd position from right. In 4th step, the 2nd smallest number (15) comes at the 2nd position from left and 26 get arranged automatically coming at 3rd position from left.

Note:
Left-right (or right-left) arrangement of numbers also take place in the same manner when numbers are arranged in decreasing order.

Arrangement of Words and Numbers Simultaneously:
Just see the following outputs produced by a word and number machine.

Case I

In such case, numbers and words get arranged alternately. In step I, the smallest number (32) comes at the 1 st position from left pushing the remaining members of input towards right. In the step II, the word coming 1st alphabetically (that is the word ‘ Aradhna’) takes the 2nd position from left pushing the remaining member rightward. Point to be noted that the 2nd smallest number automatically comes at the third position from left while arranging the word ‘Aradhna’ and hence, there is no need to arrange the 2nd smallest number ‘50’. In step III, the word (Prema) coming 2nd alphabetically comes at the 4th position from left pushing the other members to the right. In step IV, the largest number (100) occupies the 5th position from left and the word (Vandana) coming last alphabetically comes at last position automatically finishing the complete arrangement.

Let us see some other cases of this type:

Case II:

In this case, largest number and the word coming last alphabetically get arranged alternately. Then the 2nd longest number and the word coming 2nd last alphabetically get arranged alternately and the process goes on till the arrangements of all the numbers and words get completed. In this case, arrangement completes in step III.

Case lll:

In this case, arrangement starts with the word coming 1 st alphabetically and such word is ‘ Aradhna’ that comes at the 1 st position from left is step 1. In step
II, the smallest number (32) comes at the 2nd position from left. Then, in step
III, the word coming 2nd alphabetically comes at the 3rd position from left and all the other members get arranged automatically.

Case IV:

In this case, word coming last alphabetically comes 1 st from left in step I and such word is ‘Vandana’. In step II, the largest number (100) comes at the 2nd position from left. In step III, the word coming 2nd last alphabetically occupies the 3rd position from left, and such word is ‘Prerna’. As the 2nd largest number (50) automatically get arranged as per the pattern going on and hence this is not needed to arranged in step IV. In step VI, the word coming 1st alphabetically comes at the 5 th position from left and such word is ‘Aradhna’. The smallest number (32) get arranged automatically coming at the last position from left in step IV. Thus, it is clear that in this case the word coming 1st alphabetically and the greatest number get arranged alternately in 1 st two steps; then 2nd last word alphabetically and 2nd largest number get arranged alternately finishing the whole arrangement in step IV.

CaseV:

In this case, the smallest number comes at the 1st position from left in step I and such number is 32. In step II, the word (Vandana) coming last alphabetically occupies the 2nd place from left. In the 2nd step, the 2nd smallest number (50) takes the 3rd position from left automatically and also the word coming 2nd last alphabatically takes the 4th position from left automatically. Hence, there is no need to arrange ‘50’ and ‘Prema’. In the III step, the largest number (100) occupies the 5th position from left completing the whole arrangement.

Case VI:

In this case, the logic is that the greatest number (100) comes at the 1 st position from left in step I. In step II the word coming 1st alphabetically takes the 2nd position from left and the 2nd largest number (50) gets arranged automatically. Hence, in step III, we direct arrange the word coming 2nd last alphabetically (that word is ‘Prerna’) occupies the 4th position from left and the other two members (32 and ‘Vandana’) get arranged automatically finishing the whole arrangement.

Arrangement Based on the Number of Letters in Words:
Just have a look at the following patterns:

Case I:

Here, the words get arranged as per increasing number of letters.

words, the word having least number of letters comes 1 st from left in step I and such word is ‘be’. The word ‘let’ is bigger than ‘be’ and smaller than other words letterwise and hence, it takes 2nd position from left but it gets arranged automatically when the word ‘be’ is arranged in step I. In 2nd step, the word ‘love’ comes at the 3rd position from left as it is bigger than word ‘let’ letterwise. In step III, the letterwise bigger word (fried) than love comes at the fourth position from left. Similarly, mature comes at the 5th position from left and pattern comes at the last position automatically while arranging the word ‘mature’.

Case II:

In this case, the words get arranged in decreasing order in terms of letters. In other words, the word having the largest number of letters comes 1 st from left, then comes the word having 2nd largest number of letters, then comes the word having 3rd largest number of letters and the process goes on till the word having the least number of letters occupies the last position from left.

Case III:

Have you noticed something here? Here, the words get arranged in increasing order of litters. But when it comes to the case of two or more words having equal number of letters the priority is given alphabetically. It does mean that the word coming 1 st as per the alphabet will be put before the word coming 2nd. Similarly, the word coming 2nd alphabetically will be put before the word coming third. This is the reason why ‘let’ has been put before ‘set’ and ‘gate’ has been put before ‘hope’.

Case IV:

In this case, the words get arranged in decreasing order of letters. But when it comes to the case of two or more words having equal number of letters the priority is given to the word that comes later alphabetically. It does mean that the word coming 1 st alphabetically will be put after the word coming 2nd and the word coming 2nd will be put after the word coming 3rd. This is the reason why ‘hope’ has been put before ‘gate’ and ‘set’ has been put before ‘let’.

Important Note: The case ofarrangement discussed so far are the cases ofpush. In all the cases a new word jumps from its place in every step, occupies its new and due place and gives the remaining words and push either towards lej’t or right as per the requirement of the pattern. But in some cases of arrangement interchange does take place and that format is given below:

Arrangement with Interchange:
EXAMPLE:

In this case, the word (beautiful) coming 1 st in alphabetical order comes at the 1st position from left interchanging its place with the word ‘the’ and this is step I. In step II, the word (girl) coming 2nd in alphabetical order occupies the 2nd position from left interchanging with the word ‘most’. In step III, the word coming 3rd (is) comes at the third position from left interchanging with the word ‘the’ and finishing the complete arrangement in alphabetical order.
This type of cases can also be seen in number arrangements and in the arrangements of numbers and words simultaneously. The examples of these type of arrangements are given below:

EXAMPLE (Increasing order number arrangement)

Presentation:

The presentation gives you the clear idea of how interchange takes place in every step.

EXAMPLE (Decreasing order number arrangement)

Presentation:

PROBLEMS OF MATHEMATICAL OPERATION—
In this type of problems, the input has some numbers. Different steps are obtained by taking the numbers of the input and dilferent arithmetic operations are performed after that.

EXAMPLE:

In this case, in step I (each number of the input – 8). In step II, product of the digits of each number of the input. In step III, each number of the input is multiplied by 3. In step IV, each number of the input is added by 6. In step V, keep adding the digits of each number of the input till they are converted into single digit. In step VI, (digit sum of each number of input) . In step VII, each number of step II is added by 4.

MISCELLANEOUS PROBLEMS

In this type of problems, there is no fixed pattern of questions coming under this category. Infact, questions under this category comes before you as a real surprise.

In this case, in step I, first letter disappear. In step II, last letter disappear. In step III, vowels disappear. In step IV, consonants disappear. In step V, first two letters disappear.

Shortcut Approach

• First of all, observe the given input line of words or numbers and the last step of rearrangement, so that candidate may get an idea about the changes effected in various steps of rearrangement.
• In order to know what changes have been made in each step, observe two consecutive steps carefully.
• Now, correlate the input, the last step and anyone of the middle steps. This will enable you to identify the rule of arrangement.
• In shifting problems, it is possible to determine the previous/earlier steps including input. We can proceed/move backward or in reverse direction in shifting problems.
• In shifting problems for convenience, we assign numeric value to given words.

Shortcuts in Reasoning for Competitive Exams – Cube & Dice

Shortcuts in Reasoning Quantitative Aptitude English

CUBE

INTRODUCTION

• A cube is three dimensional figure whose length, breadth and height are equal
  and any two adjacent faces are inclined to each other at 90°. It has 6 faces, 8 corners and 12 edges.
  
• Corners of the cube are A, B, C, D, E,F,G and H
• Edges of the cube are AB, BE, EF, AF, AD, CD, BC, EH, CH, GH, DG and FG.
• Faces of the cube are ABCD, EFGH, CDGH, BCHE, ABEF and ADFG.

When a cube is painted on all of its faces with any colour and further divided into various smaller cubes of equal size, we get following results:

• Smaller cubes with no face painted will present inside faces of the undivided cube.
• Smaller cubes with one face painted will present on the faces of the undivided cube.
• Smaller cubes with two faces painted will present on the edges of undivided cube.
• Smaller cubes with three faces painted will present on the comers of the undivided cube.

The above figure may be analysed by dividing it into three horizontal layers :

Layer I or top layer:

The central cube has only one face coloured, four cubes at the corner have three faces coloured and the remaining 4 cubes have two faces coloured.

 

Layer II or middle layer:

The central cube has no face coloured, the four cubes at the corner have two faces coloured and the remaining 4 cubes have only face coloured.

 

 

Layer III or bottom layer:

The central cube has only one face coloured, four cubes at the corner have three faces coloured and the remaining 4 cubes have two faces coloured.

Also, if
n = no. of divisions on the faces of cube

Shortcut Approach

• Number of smaller cubes with no face painted = (n – 2)³
• Number of smaller cubes with one face painted = (n – 2) * 6
• Number of smaller cubes with two faces painted = (n – 2) * 12
• Number of smaller cubes with three faces painted = 8

EXAMPLE

 A cube is painted blue on all faces is cut into 125 cubes of equal size.
Now, answer the following  question?

How many cubes are not painted on any face?

Solution:-  Since, there are 25 smaller cubes of equal size, therefore

n = number of divisions on the face of undivided cube = 5.
Number of cubes with no face painted = (n – 2)³
= (5-2)³ = 27

DICE

INTRODUCTION

A dice is three-dimensional figure with 6 surfaces. It may be in the form of a cube or a cuboid. After observing these figures, we have to find the different side (opposite or adjacent sides) of the dice.

Dice Formation

A Dice is formed by folding a sheet of paper. These forms may be

Form 1:
Number 1 is opposite to 5.
Number 2 is opposite to 4.
Number 3 is opposite to 6.

 

Form 2:
Number 1 is opposite to 6.
Number 2 is opposite to 4.
Number 3 is opposite to 5.

Form 3:
Number 1 is opposite to 3.
Number 2 is opposite to 5.
Number 4 is opposite to 6.

Form 4:
Number 1 is opposite to 4.
Number 2 is opposite to 6.
Number 3 is opposite to 5.

Form 5:
Number 1 is opposite to 4.
Number 2 is opposite to 5.
Number 3 is opposite to 6.

1. Ordinary Dice:
In this type of dice, the sum of opposite sides is not 7 but the sum of two adjacent sides are seven.

 

2. Standard Dice:
In such type of dice, the sum of opposite sides is 7 or sum of adjacent side is not 7.

 

Opposite of 1……. 6 (since 1+6 =7)
Opposite of 5……. 2 (since 5+2 =7)
Opposite of 3……. 4 (since 3+4 =7)

IMPORTANT RULES

Rule-1 :

If two sides of cubes are common( has same numbers or symbols), then the remaining two will be opposites of each other.

In above shown two dices, number 4 and 5 are
common in both dices, hence, 3 and 1
will be apposite to each other.

Rule 2:

If one side of dices is common

If one side of given dices are common then list these sides (numbers on them) either in clock-wise or anti-clockwise. Comparing the numbers obtained from both dices will give you the opposite numbers.

In this figure, number 2  is common in both dices. Now, writing the remaining no, in
clock-wise direction, we get:

2…………..1………….. 4 (dice 1)
2………….. 3………….. 6(dice 2)
Through the above observed data, we can say that:
1 is opposite to 3
4 is opposite to 6
2 is opposite to 5

Rule 3 :

If one side is common and it’s place is same in both dices.

If one side is common in both cubes and it’s place is same in both of these dices, then the remaining two sides of respective dices which appear in figure will be the opposite of each other.

 

As you can see,number 2 is common in both of these dices and it appears in the same face in both these dices. In such case, the remaining two sides in both dices will be opposite to each other.

In this figure, the opposite sides are :

• 4 is opposite to 3 (as the position of 4 and 3 are same on two dices)
• 6 is opposite to 1 (as the position of 6 and 1 are same on two dices)
• 2 is opposite to 5 (we already know the position of 1,6,3,4 and 2.
  The only one remaining is 5)

EXAMPLE

Two positions of a dice are shown, when 4 is at the bottom,
what number will be on the top?

Solution:-
From the two figures it is clear that the numbers 2, 3, 5 and 6 cannot appear opposite 1.
So, 4 appears opposite 1. Therefore, when 4 is at the bottom, 1 will be on the top.

Shortcuts in Reasoning for Competitive Exams – Analytical Decision Making

Shortcuts in Reasoning Quantitative Aptitude English

Analytical Decision Making is based on a set of relationships laid out, generally arbitrarily, from which new information can be deduced. This involves two steps-first of analysis and second of reasoning. Analytical decision making deals with questions in which you have to decide upon the course of action taken upon a candidate who has applied for a post or membership to an institution keeping in mind the essential requisites and the data given for the candidate.

CATEGORIES OF ANALYTICAL DECISION MAKING

Category I
In this type a vacancy is being declared. The necessary qualifications required by the recruiting agencies are given with certain exceptions. The qualifications and the merits of the candidates are mentioned. The decision about each candidate has to be made from amongst the five choices given, which state the courses of action to be taken as per the candidate’s potential.

Category II
Here, the eligibility conditions for joining a course or availing certain benefits etc are given as against the vacancies mentioned in the former category. The qualifications of the candidates are also mentioned. The decision about each candidate is to be made from among the five answer choices given.

FORMAT OF THE QUESTION

Example (Directions): Read carefully the information given below and answer the questions based on it:
The following are the given conditions for the recruitment of a candidate as a family member in a computer institute:

1. The candidate must be in the age range of 23 years to 28 years as on 1st November, 2013.
2. The candidate must have work experience as a teacher or programming experience of at least 2years.
3. The candidate must have a PG degree in computer application, [M.C.A, M.Tech. or  M.Sc.
   (computer science)] with not less than 60% marks.
4. Out of total 50 marks in the interview, the candidate must obtain 50%. In the case when a candidate
5. Fulfils the above conditions, he/she shall be appointed as senior teacher.
6. Has less than 60% but more than 50% marks in his/her PG degree in computer application, he/she will be appointed as junior teacher.
7. Is of age more than 28 years but less than 32 years as on 1st November, 2013, the case may be reffered to the GM of the institute.

On the basis of the above mentioned conditions and information about each of the candidates in the question below, you have to decide which of the following courses of action should be than against each candidate. Point to be noted that nothing extra will be assumed except the given information. The decision must be based only on the data provided.

Mark your answer:

1. If the candidate is to be selected as a Junior teacher
2. If the candidate is to be selected as a Senior teacher
3. If the case will be reffered to the GM of the institute.
4. If the data are inadequate
5. If the candidate is not to be selected.

QUESTIONS:

1. Mukesh Verma was born on 31st July, 1985. He is an M.Tech. in computer engineering with 70% marks. He has been working in an institution as a programmer for the last 7 years.
2. Karishma Tiwari is MCA with 72% marks. Her date ofbirth is 14th August, 1990. She has worked as a computer teacher for 4 years. She has got 35 marks in interview.

What You See in the given Question Format?

In the given format you can see the following things:

1. Information about some candidates have been provided.
2. Some conditions have been given for candidates to fulfill in order to get selected for a particular job/ post. In case of the given format, four conditions have been given.
3. When a candidate fulfills all the criteria except some, then different course of action has to be taken for him.

Some more things to understand

• Basic conditions: In the given question format, there are four basic conditions — (i), (ii), (iii) and (iv). They are called basic conditions because they are the original conditions.
• Additional conditions: In the given question format, there are two more conditions apart from the basic conditions and they are (vi) and (vii). point to be noted that (v) will not be on additional condition as t does not talk of exceptions. In fact (v) is only a totality of the four basic or original conditions given in the question format.

What is data inadequacy?

As one of the answer is given as ‘data inadequate’ we must be clear about what exactly does data inadequacy mean? When details given about any candidate provide no information as required by the basic conditions/additional conditions then this would be the case of data inadequacy, For example, let us see the first question given in the format. No information is given about what marks have been obtained by Mukesh Verma in the interview. Hence, the data is inadequate here.

How to solve a given problem?

Let us consider the questions given in the format and start one step wise process.

STEP I

Write the name of the candidates in the left side and then write the symbols (i, ii, iii, iv) of the basic conditions to the top right.

Now, put the symbols of the additional conditions (vi and vii) below the symbols of that basic condition with which these might be related.
For example, (vi) is a condition about educational qualification and so, it is an exception of (iii).
Hence (vi) should be written below (iii). Similarly, (vii) should be written below (i).
Now, after the completion of step I, the following format will be prepared:

		I (vii)	ii	iii (vi)	iv
1	Mukesh Verma
2	Karishma Tiwari
3	Brijesh Shankar
4	Mansi Ranjan
5	Subodh Sexena

NOTE : To differentiate between basic conditions and additional conditions.
The additional conditions have been encircled.

STEP II

At the 2nd step just see the given answer 1 choices carefully and decide which combination of the conditions leads to which conclusion. If we see the given question format with serious eye, we find that the following combination can be formed.
i + ii + iii + i v —> 2 [Senior teacher] vii + ii + iii + iv —> 3 [Case will be reffered to GM]
i + ii + vi + iv -> 1 [Junior teacher] When we have decided the above three combination giving answer choices remain and the answer choice (a), (b) and (c), two answer choices remains and they are answer choice (d) and answer choice (e). The answer choice (e), which says that the candidate is not to be selected, should be chosen when any one or more of the given conditions is violated. The answer choice (d), which tells that the data are inadequate, should be chosen when no information is given about any one or more conditions.

How to examine data?

After step II you are required to read all the statements carefully. Just take each question one by one and compare then with the given conditions. Examinees are suggested to use following symbols while doing this comparision:

1. If a basic condition is fulfilled mark ‘✓’ sign below it.
2. If a basic condition is violated and it is not attached with an additional condition then mark ‘X’ sign below it.
3. If a basic condition is violated but it is attached with an additional condition, then
   • Mark a ‘x’ sign below it if additional condition is also violated.
   • Mark a ‘✓’ sign below it if additional condition is fulfilled.
4. In case of unavailability of any information about any condition, a mark’?’ Will be put below that condition.

To understand point (i) to point (iv) let us see the presentation given below:

1. I, II, III and IV are basic conditions while (V) and (VI) are two additional conditions. (V) is attached to II and (VI) is attached to III.
2. In question (1), I, II and III are satisfied while VI is violated
3. In question (2), all the basic conditions I, II, III and IV are satisfied

STEP III

1. One by one, read the questions very carefully and compare the facts given with the various condition.
2. Mark the appropriate sign or ‘x’ or ‘ x ’ (✓) or (x)? As required
3. When a ‘x’ or a (x) sign is obtained, then stop examining further and without any hesitation select the answer choice “not to be selected” for that particular question. In another words whenever you get ‘x’ or (x) sign, do not take any botheration to examine the remaining condition, select your answer as “not to be selected and quickly move on to the next question. It so happens because,
   if a condition as well as its additional condition is violated, it does mean that one necessary requirement is not being fulfilled. Hence, we reach at a conclusion that the selection is not possible even it other conditions are fulfilled.

STEP IV

Now, this is the time to select your answer choices on the pattern given below:
If find a ‘x’ or (x) below any condition, go for the answer choice “not to be selected”
If you find no cross mark but there is a question mark below any condition, your answer choice would be “data are inadequate”, (iii) If you find neither any cross mark nor any question mark, than compare the combination with the three answer combinations obtained in step II and select the answer choice accordingly. After understanding the above steps, now we are at a position of solving the question given in the question format. Let us see the solution:
Solution:

Condition (V) is attached to II while the additional condition is VI attached with the basic condition III.

STEP WISE EXPLANATION OF ABOVE TABLE:

Step I

At the step I level, we read the question carefully and find out that there are four, basic conditions (i), (ii), (iii) and (iv) and two additional conditions (vii) and (vi). further, it is clear that ‘(vii)’ is an exception of ‘(i)’ and ‘(vi)’ is an exception of‘(iii)’. Now we write the name of the candidates in extreme left and then put the basic conditions (i), (ii), (iii) and (iv) at the top-right of the candidate in question 1.
Next, we write additional condition ‘(vii)’ below ‘o’ and additional condition ‘(vi)’ below ‘(iii)’.

Step II

At the 2nd level, we look at the answer choices and prepare one answer combinations accordingly.
This will be:
i + ii + iii + iv ⇒ b
vii + ii + iii + iv ⇒ c
i + ii + vi + iv ⇒ a

Step III

At the step III level, we read every question carefully and compare the facts given in it with the various conditions.
Let us see the detailed analysis of every candidate question wise.

Mukesh Verma
He is an M.Tech in computer engineering with 70% marks. This fulfills condition C. Hence we write ‘ ✓’ mark below C.
Next, his date of birth is 31st July, 1985. Here, we do a mental calculation that on 31st July, 2013 he turned 28th. This is the reason that on 1st November 2013, he is more than 28 years. Therefore, (i) is violated, but the additional condition of (i) is (vii) which is fulfilled and we write (✓) mark here. Further, Mukesh Verma is having a programming experience of 7 years (more than 2 years). So we mark (✓) below (ii). Lastly, there is no information about marks of Mukesh in the interview. Thus the sign of question mark ‘?’ is put below d.

Karishma Tiwari
Karishma is an MCA with 72% marks. This fulfills (iii), so we put the mark ‘✓’ below (iii). Her date of birth is 14th August, 1990, Soon 1st November, 2013, she is more than 23 years. This fulfills ‘(i)’ and hence we put a (✓) mark below ‘(i)’. She is a computer teacher from last 4 years. This fulfills (ii) and we put (✓) mark below (ii), lastly, she has obtained 35 marks in the interview. This marks is more than the required 50% (25 marks out of 50 marks), therefore (iv) is also fulfilled and we put (✓) mark below (iv).

Step IV

At 4th level we select the answer choices.
Solution 1. No cross mark. But a question mark is available. Hence, data is inadequate.
Solution 2. i + ii + iii + iv => b [step II]
So, the candidate is to be selected as a senior teacher.

Shortcut Approach

• For selection all basic conditions must be fulfilled.
• For rejection atleast one independent basic condition must I be violated/basic ‘+’ additional condition must be violated.
• If a basic condition is violated but an additional condition attached with it is fulfilled and all other,remaining basic conditions are fulfilled, then the case will be referred to the person given in the questions.
• Once the symbol x/(x) is put in the table, there is no need to check further conditions as person is . declared rejected at this stage only.
• If for one basic condition, the data is not given while all other basic, conditions are fulfilled, it means data is inadequate. .
• If any information is not given, I and answer choices don’t have] data inadequate option, then, condition related to that particular information is supposed to be violated.